Let $g(x)=\begin{cases} \dfrac{1}{\cos(x)}&\text{for }-\dfrac{\pi}{2}<x < 0 \\\\ \cos(x+\pi)&\text{for }x\geq 0 \end{cases}$ Is $g$ continuous at $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: For $g$ to be continuous at $x=0$, we need $\lim_{x\to 0}g(x)$ and $g(0)$ to exist and be equal. Since $0\geq 0$, the rule that applies to $x=0$ is $\cos(x+\pi)$. So $g(0)=\cos(0+\pi)=-1$. Now let's analyze $\lim_{x\to 0}g(x)$. Finding $\lim_{x\to 0^{ +}}g(x)$ For $x$ -values larger than $0$, the appropriate rule for $g(x)$ is $\cos(x+\pi)$. Since $\cos(x+\pi)$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ +}}g(x) \\\\ &=\lim_{x\to 0^{ +}}[\cos(x+\pi)] \gray{\cos(x+\pi)\text{ is the rule for }x>0} \\\\ &=\cos(0+\pi) \gray{\cos(x+\pi)\text{ is continuous at }x=0} \\\\ &=-1 \end{aligned}$ Finding $\lim_{x\to 0^{ -}}g(x)$ For $x$ -values smaller than $0$, the appropriate rule for $g(x)$ is $\dfrac{1}{\cos(x)}$. Since $\dfrac{1}{\cos(x)}$ is continuous for $-\dfrac{\pi}{2}<x < 0$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ -}}g(x) \\\\ &=\lim_{x\to 0^{ -}}\dfrac{1}{\cos(x)} \gray{\dfrac{1}{\cos(x)}\text{ is the rule for }x<0} \\\\ &=\dfrac{1}{\cos(0)} \gray{\dfrac{1}{\cos(x)}\text{ is continuous at }x=0} \\\\ &=1 \end{aligned}$ Conclusion We found that $\lim_{x\to 0^{ +}}g(x)=-1$ and $\lim_{x\to 0^{ -}}g(x)=1$. Since the one-sided limits aren't equal, $\lim_{x\to 0}g(x)$ doesn't exist and $g$ isn't continuous at $x=0$.